Hard to buy gifts [half]

Description–

Small X students to the students chose C N small gift, decided to buy and gift order, but as a poor child did not pay no pocket money, there are M-bit intentioned students lent a helping hand, however, in order to reduce the highest borrowing volume, small X students OI hope you contest his rational planning, so that he can be relaxed and happy to send gifts.

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Input–

The first input line separated by a space of two positive integers N and M are
the following N lines each a positive integer not more than 10,000, the gift order Prices.

Output–

An integer that is the maximum loan amount.

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Sample Input–

7 5
100
400
300
100
500
101
400

Sample Output–

500

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Notes -

30%：n <=10
60%: n<=1000
100%: n<=100000

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Code -

#include<iostream>
#include<cstdio>
using namespace std;
int n,m,l,r,s,t,fy,mid,a[100005];
int main()
{
	scanf("%d%d",&n,&m);
	for (int i=1;i<=n;++i)
	{
		scanf("%d",&a[i]);
		l=max(l,a[i]),r+=a[i];
	}
	while(l<=r)//二分答案
	{
		mid=(l+r)/2;
		s=mid-a[1];
		t=m,fy=0;
		for (int i=2;i<=n;++i)
		 if (s>=a[i]) s-=a[i];
		 else if(t>0) s=mid-a[i],t--;
		      else { fy=1; break; }
		if (fy) l=mid+1;
		else r=mid-1;
	}
	printf("%d",l);
	
	return 0;
}